Nguyên hàm (2x-1)ln(x+2))

$\quad \displaystyle\int(2x-1)\ln(x+2)dx$

Đặt $\begin{cases}u = \ln(x+2)\\dv = (2x-1)dx\end{cases}\longrightarrow \begin{cases}du =\dfrac{1}{x+2}dx\\v =\dfrac14(2x-1)^2\end{cases}$

Ta được:

$\dfrac14(2x-1)^2\ln(x+2) -\dfrac14\displaystyle\int \dfrac{(2x-1)^2}{x+2}dx$

$= \dfrac14(2x-1)^2\ln(x+2) -\dfrac14\displaystyle\int\left(4x +\dfrac{25}{x+2} -12\right)dx$

$= \dfrac14(2x-1)^2\ln(x+2) -\dfrac14\left(4\displaystyle\int xdx + 25\displaystyle\int\dfrac{1}{x+2}dx - 12\displaystyle\int dx\right)$

$= \dfrac14(2x-1)^2\ln(x+2) - \dfrac{x^2}{2} - \dfrac{25}{4}\ln|x+2| - 3x + C$